Question

For the game of interstellar baseball, teams have used robotic enhancements to allow superhuman feats. Consider a pitcher who can throw a baseball at 65 % of the speed of light and is practicing in a spaceship traveling away from Earth at 95 % of the speed of light (on the way to the team's next game).

Assuming that he throws the baseball in the same direction the spacecraft is traveling, how fast will those of us on Earth measure the baseball to be going?

Answer 1

98.91 % of the speed of light as seen from the Earth

Step-by-step explanation:

This problem involves the relativistic effects. The pitcher throws the ball with a velocity u​​​​​​​​​​​​' which is 60% of the speed of light c i.e. u'=0.65c

The velocity V of the spaceship in which he is practicing is 95% of the speed of light. Therefore V=0.95c

Therefore, if the ball is thrown in the same direction as the direction of the spacecraft's motion,then typically, the velocities would add up and exceed the speed of light. Butthis is not possible. It is prohibited by the special theory of relativity. So we need to consider the relativistic addition of velocities to calculate the velocity (u) of the ball as seen from the Earth.

The formula for relativistic addition of velocities is expressed as follows

1+u'/(1+Vu'/c^2)

Given that u'=0.65c and V=0.95c. we can calculate u which is the velocity of the baseball as seen from Earth as follows

= 0.95c+ 0.65c/(1+ 0.95c×0.65c/c^2)

1.60c/(1+0.6175c^2/c^2)

=0.9891c

In conclusion, the baseball would seem to be traveling with a velocity 0.9891c of 98.91 % of the speed of light as seen from the Earth .

Answer 2

The baseball would seem to be traveling with a velocity 0.98918c of 98.92% of the speed of light as seen from the Earth.

Step-by-step explanation:

This problem involves the relativistic effects. The pitcher throws the ball with a velocity u​​​​​​​​​​​​' which is 65% of the speed of light c i.e. u' = 0.65c

The velocity V of the spaceship in which he is practicing is 95% of the speed of light. Therefore V = 0.95c

So if the ball is thrown in the same direction as the direction of the spacecraft's motion, then classically, the velocities would add up and exceed the speed of light. But we know that this is not possible. It is prohibited by the special theory of relativity. So we need to consider the relativistic addition of velocities to calculate the velocity (u) of the ball as seen from the Earth.

The formula for relativistic addition of velocities is expressed below,

`u=(V+u')/(1+(Vu')/(c^2))`

We know that u' = 0.65c and V = 0.95c. now we can calculate u which is the velocity of the baseball as seen from Earth as follows

`u=(0.95c+0.65c)/(1+(0.95c* 0.65c)/(c^2))`

`u=(1.6c)/(1+(0.6175c^2)/(c^2))`

`u = (1.6c)/(1.6175)`

u = 0.98918c

So the baseball would seem to be traveling with a velocity 0.98918c of 98.92% of the speed of light as seen from the Earth.